Quantum Mechanics

Part III. Quantum Bootstrap*

Harmonic Oscillator

Position and Momentum

Discrete v.s. Continuous

Examples: position x of a quantum particle.

where

(OverHat[x]) denotes the position operator (a Hermitian operator corresponding to the position observable of the particle)

x∈ℝ is the position eigenvalue.

(Ket[x]) the corresponding position eigenstate (the quantum state that describe the particle at the position x).

The Hilbert space dimension is infinite. It is always helpful to think about the continuous eigen spectrum as the limit of an infinitely dense discrete spectrum.

({{ , discrete, →, continuous}, {orthonormal basis, (BraKet[i, j])=(Subscript[δ, i j]), →, (BraKet[x, x'])=δ(x-x')}, {resolution of identity, (∑_i)(Ket[i])(Bra[i])=𝟙, →, ∫dx(Ket[x])(Bra[x])=𝟙}, {state decomposition, (Ket[v])=(∑_i)(Subscript[v, i])(Ket[i]), →, (Ket[ψ])=∫dx ψ(x)(Ket[x])}, { , (Subscript[v, i])=(BraKet[i, v]),  , ψ(x)=(BraKet[x, ψ])}, {state normalization, (∑_i)((|(Subscript[v, i])|)^2)=1, →, ∫dx((|ψ(x)|)^2)=1}, {scalar product, (BraKet[u, v])=(∑_i)(BraKet[u, i])(BraKet[i, v])
=(∑_i)(Subscript[u, i]^( * ))(Subscript[v, i]), →, (BraKet[ϕ, ψ])=∫dx(BraKet[ϕ, x])(BraKet[x, ψ])
=∫dx (SuperStar[ϕ(x)])ψ(x)}})

Dirac delta function - the continuous limit of the Kronecker delta symbol. It is defined by the following property under integration

∀f:∫dx δ(x)f(x)=f(0).

Position Operator

The position operator is diagonal in its own eigen basis.

Suppose the particle is in a state (Ket[ψ])

(Ket[ψ])=∫dx ψ(x)(Ket[x]),

described by the wave function ψ(x).

Applying the position operator,

(OverHat[x])(Ket[ψ])=∫dx (x ψ(x))(Ket[x]).

Translation

Suppose the particle was in the (Ket[x]) state (at position x).

After applying the translation operator, the particle is in a new state (Ket[x+a]) (at position x+a).

Therefore (OverHat[T])(a) translates the particle by displacement a.

Translation operator implements a basis transformation (from (Ket[x]) to (Ket[x+a])). Every basis transformation is unitary. So the translation operator is unitary.

Use the definition Eq. () to show that 
(SuperDagger[(OverHat[T])(a)])(OverHat[T])(a)=(OverHat[T])(a)(SuperDagger[(OverHat[T])(a)])=𝟙, 
thus the translation operator is unitary.

(SuperDagger[(OverHat[T])(a)])=∫dx (Ket[x])(Bra[x+a]).

(SuperDagger[(OverHat[T])(a)])(OverHat[T])(a)=∫dx' (Ket[x'])(Bra[x'+a])∫dx (Ket[x+a])(Bra[x])
=∫dx'∫dx (Ket[x'])(BraKet[x'+a, x+a])(Bra[x])
=∫dx'∫dx (Ket[x'])δ(x'+a-x-a)(Bra[x])
=∫dx'∫dx (Ket[x'])δ(x'-x)(Bra[x])
=∫dx (Ket[x])(Bra[x])
=𝟙.

Momentum Operator

(OverHat[p])=i ℏ(∂_a)(OverHat[T])(a)(Subscript[|, a=0]),
=i ℏ (lim_(a→0))(((OverHat[T])(a)-(OverHat[T])(0))/a),

Suppose the particle is in a state (Ket[ψ])

(Ket[ψ])=∫dx ψ(x)(Ket[x]),

described by the wave function ψ(x).

Under translation,

(OverHat[T])(a)(Ket[ψ])=∫dx ψ(x)(Ket[x+a])
=∫dx ψ(x-a)(Ket[x]).

Applying the momentum operator,

(OverHat[p])(Ket[ψ])=i ℏ (lim_(a→0))(((OverHat[T])(a)(Ket[ψ])-(OverHat[T])(0)(Ket[ψ]))/a)
=i ℏ∫dx ( (lim_(a→0))((ψ(x-a)- ψ(x))/a))(Ket[x])
=∫dx ( -i ℏ(∂_x)ψ(x))(Ket[x]).

Show that Eq. () is consistent with Eq. () when acting on a state (Ket[ψ]).

(OverHat[p])(Ket[ψ])=-i ℏ∫dxdx'(Ket[x])(∂_x)δ(x-x')(Bra[x'])∫dx'' ψ(x'')(Ket[x''])
=-i ℏ∫dxdx'dx''(Ket[x])(∂_x)δ(x-x')ψ(x'')(BraKet[x', x''])
=-i ℏ∫dxdx'dx''(Ket[x])(∂_x)δ(x-x')ψ(x'')δ(x'-x'')
=-i ℏ∫dxdx'(Ket[x])(∂_x)δ(x-x')ψ(x')
=-i ℏ∫dx(Ket[x])(∂_x)ψ(x)
=∫dx (-i ℏ(∂_x)ψ(x))(Ket[x]).

Commutation Relation

(OverHat[x])≏x, (OverHat[p])≏-i ℏ(∂_x),

[(OverHat[x]),(OverHat[p])](Ket[ψ])≏[x,-i ℏ(∂_x)]ψ(x)
=-i ℏ(x(∂_x)-(∂_x)x)ψ(x)
=i ℏ ψ(x)
≏i ℏ(Ket[ψ]).

Operator Algebra

Hamiltonian

(OverHat[H])=(1/(2m))((OverHat[p])^2)+(1/2)m (ω^2)((OverHat[x])^2).

[(OverHat[x]),(OverHat[p])]=i ℏ.

(OverHat[p])->(OverHat[p])(Sqrt[ℏ m ω]), (OverHat[x])->(OverHat[x])(Sqrt[(ℏ/(m ω))]),

Energy scale set by ℏ ω.

New operators (OverHat[x]) and (OverHat[p]) are dimensionless.

Commutation relation for the rescaled operators

Annihilation and Creation Operators

(OverHat[a]) and (SuperDagger[(OverHat[a])]) are Hermitian conjugate to each other.

Analogy: complex numbers z=x+i y, (SuperStar[z])=x-i y => position (OverHat[x]) ∼ real part of (OverHat[a]), momentum (OverHat[p]) ∼ imaginary part of (OverHat[a]).

(OverHat[a]) (SuperDagger[(OverHat[a])])=(SuperDagger[(OverHat[a])])(OverHat[a])+𝟙.

((OverHat[a])^l) ((((SuperDagger[(OverHat[a])])))^m)=(∑_(k=0)^(min(m,l)))((m!l!)/((m-k)!(l-k)!k!))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l-k))

Prove Eq. ().

Base case: (l,m)=(1,1),

(OverHat[a]) (SuperDagger[(OverHat[a])])=(∑_(k=0)^1)((1!1!)/((1-k)!(1-k)!k!))((((SuperDagger[(OverHat[a])])))^(1-k))((OverHat[a])^(1-k))
=((1!1!)/((1-0)!(1-0)!0!))((((SuperDagger[(OverHat[a])])))^(1-0))((OverHat[a])^(1-0))+((1!1!)/((1-1)!(1-1)!1!))((((SuperDagger[(OverHat[a])])))^(1-1))((OverHat[a])^(1-1))
=(SuperDagger[(OverHat[a])])(OverHat[a])+𝟙,

which is true by Eq. ().

Induction along m. Assuming Eq. () is true for (l,m)=(1,m),

(OverHat[a]) ((((SuperDagger[(OverHat[a])])))^m)=(∑_(k=0)^1)((m!1!)/((m-k)!(1-k)!k!))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(1-k))
=((m!1!)/((m-0)!(1-0)!0!))((((SuperDagger[(OverHat[a])])))^(m-0))((OverHat[a])^(1-0))+((m!1!)/((m-1)!(1-1)!1!))((((SuperDagger[(OverHat[a])])))^(m-1))((OverHat[a])^(1-1))
=((((SuperDagger[(OverHat[a])])))^m)(OverHat[a])+m ((((SuperDagger[(OverHat[a])])))^(m-1)).

Consider the case of (l,m)=(1,m+1),

(OverHat[a]) ((((SuperDagger[(OverHat[a])])))^(m+1))=(OverHat[a]) ((((SuperDagger[(OverHat[a])])))^m)(SuperDagger[(OverHat[a])])
=(((((SuperDagger[(OverHat[a])])))^m)(OverHat[a])+m ((((SuperDagger[(OverHat[a])])))^(m-1)))(SuperDagger[(OverHat[a])])
=((((SuperDagger[(OverHat[a])])))^m)(OverHat[a])(SuperDagger[(OverHat[a])])+m ((((SuperDagger[(OverHat[a])])))^m)
=((((SuperDagger[(OverHat[a])])))^m)((SuperDagger[(OverHat[a])])(OverHat[a])+𝟙)+m ((((SuperDagger[(OverHat[a])])))^m)
=((((SuperDagger[(OverHat[a])])))^m)(SuperDagger[(OverHat[a])])(OverHat[a])+((((SuperDagger[(OverHat[a])])))^m)+m ((((SuperDagger[(OverHat[a])])))^m)
=((((SuperDagger[(OverHat[a])])))^(m+1))(OverHat[a])+(m+1) ((((SuperDagger[(OverHat[a])])))^m),

which is consistent with Eq. (). So Eq. () is true for any (l,m)=(1,m) with m∈(SubPlus[ℕ]).

Induction along l. Assuming Eq. () is true for (l,m), consider the case of (l+1,m),

((OverHat[a])^(l+1)) ((((SuperDagger[(OverHat[a])])))^m)=(OverHat[a])((OverHat[a])^l) ((((SuperDagger[(OverHat[a])])))^m)
=(∑_(k=0)^(min(m,l)))((m!l!)/((m-k)!(l-k)!k!))(OverHat[a]) ((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l-k))
=(∑_(k=0)^(min(m,l)))((m!l!)/((m-k)!(l-k)!k!))(((((SuperDagger[(OverHat[a])])))^(m-k))(OverHat[a])+(m-k)((((SuperDagger[(OverHat[a])])))^(m-k-1)))((OverHat[a])^(l-k))
=(∑_(k=0)^(min(m,l)))((m!l!)/((m-k)!(l-k)!k!))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k))+(∑_(k=0)^(min(m,l)))((m!l!)/((m-k)!(l-k)!k!))(m-k)((((SuperDagger[(OverHat[a])])))^(m-k-1))((OverHat[a])^(l-k))
=(∑_(k=0)^(min(m,l)))((m!l!)/((m-k)!(l-k)!k!))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k))+(∑_(k=1)^(min(m,l)+1))((m!l!)/((m-k+1)!(l-k+1)!(k-1)!))(m-k+1)((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k))
=(∑_(k=0)^(min(m,l)))((m!l!)/((m-k)!(l-k)!k!))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k))+(∑_(k=1)^(min(m,l)+1))((m!l!)/((m-k)!(l-k)!k!))(k/(l-k+1))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k))
=(∑_(k=0)^(min(m,l+1)))((m!l!)/((m-k)!(l-k)!k!))(1+(k/(l-k+1)))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k))
=(∑_(k=0)^(min(m,l+1)))((m!l!)/((m-k)!(l-k)!k!))((l+1)/(l-k+1))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k))
=(∑_(k=0)^(min(m,l+1)))((m!(l+1)!)/((m-k)!(l+1-k)!k!))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(l+1-k)),

which is consistent with Eq. (). So Eq. () is true for all l,m∈(SubPlus[ℕ]).

Corner cases:

l=0 case

((((SuperDagger[(OverHat[a])])))^m)=(∑_(k=0)^0)((m!0!)/((m-k)!(0-k)!k!))((((SuperDagger[(OverHat[a])])))^(m-k))((OverHat[a])^(0-k))
=((m!0!)/((m-0)!(0-0)!0!))((((SuperDagger[(OverHat[a])])))^(m-0))((OverHat[a])^(0-0))
=((((SuperDagger[(OverHat[a])])))^m),

which is true.

m=0 case

((OverHat[a])^l) =(∑_(k=0)^0)((0!l!)/((0-k)!(l-k)!k!))((((SuperDagger[(OverHat[a])])))^(0-k))((OverHat[a])^(l-k))
=((0!l!)/((0-0)!(l-0)!0!))((((SuperDagger[(OverHat[a])])))^(0-0))((OverHat[a])^(l-0))
=((OverHat[a])^l),

which is true.

Number Operator

(OverHat[n])=(1/2)(((OverHat[p])^2)+((OverHat[x])^2))-(𝟙/2).

Verify Eq. () using Eq. ().

(OverHat[n])=(1/2)((OverHat[x])-i (OverHat[p]))((OverHat[x])+i (OverHat[p]))
=(1/2)(((OverHat[x])^2)+i (OverHat[x])(OverHat[p])-i (OverHat[p])(OverHat[x])+((OverHat[p])^2))
=(1/2)(((OverHat[x])^2)+i [(OverHat[x]),(OverHat[p])]+((OverHat[p])^2))
=(1/2)(((OverHat[x])^2)-𝟙+((OverHat[p])^2)).

(OverHat[H])(Ket[n])=ℏ ω (n+(1/2))(Ket[n]),

(Subscript[E, n])=ℏ ω (n+(1/2)).

Quantum Bootstrap

General Principles

(OverHat[n])(Ket[n])=n(Ket[n]).

Eigen condition

This can be seen from

(Bra[n])(OverHat[n])(OverHat[O])(Ket[n])=(Bra[n])(OverHat[O])(OverHat[n])(Ket[n])=n(Bra[n])(OverHat[O])(Ket[n]).

Positivity constraint

as the squared norm of the state vector (OverHat[O])(Ket[n]) must always be non-negative.

Level Quantization

This covers the several operators as special cases,

(Subscript[(OverHat[O]), 0,0])=𝟙, 
(Subscript[(OverHat[O]), 0,1])=(OverHat[a]), 
(Subscript[(OverHat[O]), 1,0])=(SuperDagger[(OverHat[a])]), 
(Subscript[(OverHat[O]), 1,1])=(SuperDagger[(OverHat[a])])(OverHat[a])=(OverHat[n]).

The indices m,l interchange under Hermitian conjugate

(Subscript[(OverHat[O]), m,l]^†)=(Subscript[(OverHat[O]), l,m]).

Operator product expansion: product of two operators can be expanded into linear combination of operators.

(Subscript[(OverHat[O]), k,l])(Subscript[(OverHat[O]), m,n])=(∑_(p=0)^(min(m,l)))((m!l!)/((m-p)!(l-p)!p!))(Subscript[(OverHat[O]), k+m-p,l+n-p]).

Prove Eq. () using Eq. ().

(Subscript[(OverHat[O]), k,l])(Subscript[(OverHat[O]), m,n])=((((SuperDagger[(OverHat[a])])))^k)((OverHat[a])^l)((((SuperDagger[(OverHat[a])])))^m)((OverHat[a])^n)
=((((SuperDagger[(OverHat[a])])))^k)((∑_(p=0)^(min(m,l)))((m!l!)/((m-p)!(l-p)!p!))((((SuperDagger[(OverHat[a])])))^(m-p))((OverHat[a])^(l-p)))((OverHat[a])^n)
=(∑_(p=0)^(min(m,l)))((m!l!)/((m-p)!(l-p)!p!))((((SuperDagger[(OverHat[a])])))^(k+m-p))((OverHat[a])^(l+n-p))
=(∑_(p=0)^(min(m,l)))((m!l!)/((m-p)!(l-p)!p!))(Subscript[(OverHat[O]), k+m-p,l+n-p]).

(OverHat[n])(Subscript[(OverHat[O]), m,l])=(Subscript[(OverHat[O]), m+1,l+1])+m (Subscript[(OverHat[O]), m,l]),
(Subscript[(OverHat[O]), m,l])(OverHat[n])=(Subscript[(OverHat[O]), m+1,l+1])+l (Subscript[(OverHat[O]), m,l]).

[(OverHat[n]),(Subscript[(OverHat[O]), m,l])]=(OverHat[n])(Subscript[(OverHat[O]), m,l])-(Subscript[(OverHat[O]), m,l])(OverHat[n])=(m-l)(Subscript[(OverHat[O]), m,l]),

(Bra[n])[(OverHat[n]),(Subscript[(OverHat[O]), m,l])](Ket[n])=(m-l)(Bra[n])(Subscript[(OverHat[O]), m,l])(Ket[n])=0.

If m!=l (i.e. m-l≠0), for Eq. () to hold, we must have

(Bra[n])(Subscript[(OverHat[O]), m,l])(Ket[n])=0  (for m≠l).

If m=l, Eq. () is automatically satisfied. In this case, the expectation value (Bra[n])(Subscript[(OverHat[O]), m,m])(Ket[n]) can take any real number, which we denote as (Subscript[W, m|n]),

(Subscript[W, m|n]):=(Bra[n])(Subscript[(OverHat[O]), m,m])(Ket[n])∈ℝ.

(Bra[n])(Subscript[(OverHat[O]), m,m])(OverHat[n])(Ket[n])=n(Bra[n])(Subscript[(OverHat[O]), m,m])(Ket[n])=n (Subscript[W, m|n]),
(Bra[n])(Subscript[(OverHat[O]), m,m])(OverHat[n])(Ket[n])=(Bra[n])(Subscript[(OverHat[O]), m+1,m+1])(Ket[n])+m (Bra[n])(Subscript[(OverHat[O]), m,m])(Ket[n])
=(Subscript[W, m+1|n])+m (Subscript[W, m|n]).

(Subscript[W, m+1|n])=(n-m)(Subscript[W, m|n]).

(Subscript[W, 0|n])=(BraKet[n, n])=1,

(Subscript[W, m|n])=(∏_(l=0)^(m-1))(n-l).

(Bra[n]) (Subscript[(OverHat[O]), 0,m]^†)(Subscript[(OverHat[O]), 0,m])(Ket[n])=(Bra[n]) ((((SuperDagger[(OverHat[a])])))^m)((OverHat[a])^m)(Ket[n])=(Bra[n]) (Subscript[(OverHat[O]), m,m])(Ket[n])=(Subscript[W, m|n])>=0.

∀n,m:(∏_(l=0)^(m-1))(n-l)>=0

n≥0,
n (n-1)≥0,
n (n-1)(n-2)≥0,
n (n-1)(n-2)(n-3)≥0,
…

The eigenvalues n=0,1,2,… are discrete! For this reason, the operator (OverHat[n]) is called the number operator, which counts the number of elementary excitations, called phonons (the quanta of sound). The phonon is an example of emergent boson in quantum systems.

The n=0 state, denoted as (Ket[0]), is also called the vacuum state, as it describes a state with no excitations. It is also the ground state of the Hamiltonian (OverHat[H]).

The eigenstates (Ket[n]) has the following expectation value

(Bra[n])(Subscript[(OverHat[O]), m,l])(Ket[n])=(((Bra[n])((SuperDagger[(OverHat[a])])))^m)((OverHat[a])^l)(Ket[n])=({{{, ({{0, if m≠l,}, {n!/m!, if m=l.}})}})

Number Basis Representation

(OverHat[n])(Subscript[(OverHat[O]), m,l])(Ket[n])=(n+m-l)(Subscript[(OverHat[O]), m,l])(Ket[n]),

(Subscript[(OverHat[O]), m,l])(Ket[n])∝(Ket[n+m-l]).

for (Subscript[(OverHat[O]), 0,1])=(OverHat[a]),

(OverHat[a])(Ket[n])∝(Ket[n-1]);

for (Subscript[(OverHat[O]), 1,0])=(SuperDagger[(OverHat[a])]),

(SuperDagger[(OverHat[a])])(Ket[n])∝(Ket[n+1]).

(Bra[n])(SuperDagger[(OverHat[a])])(OverHat[a])(Ket[n])=(Bra[n])(OverHat[n])(Ket[n])=n,
(Bra[n])(OverHat[a])(SuperDagger[(OverHat[a])])(Ket[n])=(Bra[n])((SuperDagger[(OverHat[a])])(OverHat[a])+𝟙)(Ket[n])=(Bra[n])((OverHat[n])+𝟙)(Ket[n])=n+1.

Summary

Annihilation and creation operators

({{{, ({{(OverHat[a])=(1/(Sqrt[2]))((OverHat[x])+i (OverHat[p]))}, {(SuperDagger[(OverHat[a])])=(1/(Sqrt[2]))((OverHat[x])-i (OverHat[p]))}})}}), ({{{, ({{(OverHat[x])=(1/(Sqrt[2]))((OverHat[a])+(SuperDagger[(OverHat[a])]))}, {(OverHat[p])=(1/((Sqrt[2])i))((OverHat[a])-(SuperDagger[(OverHat[a])]))}})}}).

They satisfies the commutation relation

[(OverHat[x]),(OverHat[p])]=i 𝟙⇔[(OverHat[a]),(SuperDagger[(OverHat[a])])]=𝟙.

Number operator

(OverHat[n])=(SuperDagger[(OverHat[a])])(OverHat[a]).

It defines a discrete spectrum (OverHat[n])(Ket[n])=n(Ket[n]) for n∈ℕ. Such that

(OverHat[a])(Ket[n])=(Sqrt[n])(Ket[n-1]),
(SuperDagger[(OverHat[a])])(Ket[n])=(Sqrt[n+1])(Ket[n+1]).

Hamiltonian

(OverHat[H])=(1/2)ℏ ω (((OverHat[p])^2)+((OverHat[x])^2))=ℏ ω ((OverHat[n])+(𝟙/2)).

Eigen energies

(Subscript[E, n])=ℏ ω(n+(1/2)).

Every eigenstate (Ket[n]) can be raised from the ground state by

(Ket[n])=(1/(Sqrt[n!]))((((SuperDagger[(OverHat[a])])))^n)(Ket[0]).

Angular Momentum

Operator Algebra

Definition

(Subscript[ϵ, a b c]) is the Levi-Civita symbol: the sign of the a b c permutation.

Equivalently, in vector form, (OverHat[J])×(OverHat[J])=i (OverHat[J]).

Orbital angular momentum of a particle.

(OverHat[x])=((Subscript[(OverHat[x]), 1]),(Subscript[(OverHat[x]), 2]),(Subscript[(OverHat[x]), 3])) and (OverHat[p])=((Subscript[(OverHat[p]), 1]),(Subscript[(OverHat[p]), 2]),(Subscript[(OverHat[p]), 3])) are position and momentum operators in 3D space.

In component form, (Subscript[(OverHat[L]), a])=(Subscript[ϵ, a b c])(Subscript[(OverHat[x]), b])(Subscript[(OverHat[p]), c]).

From [(Subscript[(OverHat[x]), a]),(Subscript[(OverHat[p]), b])]=i (Subscript[δ, a b]) (set ℏ=1 for simplicity), one can verify that

[(Subscript[(OverHat[L]), a]),(Subscript[(OverHat[L]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[L]), c]).

Spin angular momentum of a qubit.

(OverHat[σ])=(((OverHat[σ])^x),((OverHat[σ])^y),((OverHat[σ])^z)) are the Pauli matrices.

The commutation relation of Pauli matrices implies

[(Subscript[(OverHat[S]), a]),(Subscript[(OverHat[S]), b])]=i (Subscript[ϵ, a b c])(Subscript[(OverHat[S]), c]).

Casimir Operator

((OverHat[J])^2) is Hermitian.

By Eq. (), one can verify that (for a=1,2,3)

[((OverHat[J])^2),(Subscript[(OverHat[J]), a])]=0.

Prove Eq. ().

[((OverHat[J])^2),(Subscript[(OverHat[J]), 3])]=[(Subscript[(OverHat[J]), 1]^2),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 2]^2),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 3]^2),(Subscript[(OverHat[J]), 3])]

[(Subscript[(OverHat[J]), 1]^2),(Subscript[(OverHat[J]), 3])]=(Subscript[(OverHat[J]), 1])[(Subscript[(OverHat[J]), 1]),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 1]),(Subscript[(OverHat[J]), 3])](Subscript[(OverHat[J]), 1])
=-i (Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 2])-i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1]),

[(Subscript[(OverHat[J]), 2]^2),(Subscript[(OverHat[J]), 3])]=(Subscript[(OverHat[J]), 2])[(Subscript[(OverHat[J]), 2]),(Subscript[(OverHat[J]), 3])]+[(Subscript[(OverHat[J]), 2]),(Subscript[(OverHat[J]), 3])](Subscript[(OverHat[J]), 2])
=i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1])+i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1]),

[(Subscript[(OverHat[J]), 3]^2),(Subscript[(OverHat[J]), 3])]=0.

Raising and Lowering Operators

In analogy to (e^(±i θ))=cos θ±i sin θ.

(Subscript[(OverHat[J]), ±]) are not Hermitian. Under Hermitian conjugate: (Subscript[(OverHat[J]), ±]^†)=(Subscript[(OverHat[J]), ∓]).

(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±]^l)=(Subscript[(OverHat[J]), ±]^l)((Subscript[(OverHat[J]), 3])±l).

(Subscript[(OverHat[J]), ∓]^(l+1))(Subscript[(OverHat[J]), ±]^(l+1))=(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±l)((Subscript[(OverHat[J]), 3])±(l+1))).

Prove Eq. ().

(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±])=(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]))
=((Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 3])+i (Subscript[(OverHat[J]), 2]))±i ((Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 3])-i (Subscript[(OverHat[J]), 1]))
=((Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]))((Subscript[(OverHat[J]), 3])±1)
=(Subscript[(OverHat[J]), ±])((Subscript[(OverHat[J]), 3])±1).

(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±]^(l+1))=(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±])(Subscript[(OverHat[J]), ±]^l)
=(Subscript[(OverHat[J]), ±])((Subscript[(OverHat[J]), 3])±1)(Subscript[(OverHat[J]), ±]^l)
=(Subscript[(OverHat[J]), ±])(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±]^l)±(Subscript[(OverHat[J]), ±])(Subscript[(OverHat[J]), ±]^l)
=(Subscript[(OverHat[J]), ±])(Subscript[(OverHat[J]), ±]^l)((Subscript[(OverHat[J]), 3])±l)±(Subscript[(OverHat[J]), ±])(Subscript[(OverHat[J]), ±]^l)
=(Subscript[(OverHat[J]), ±]^(l+1))((Subscript[(OverHat[J]), 3])±(l+1)).

Prove Eq. ().

(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])=((Subscript[(OverHat[J]), 1])∓i (Subscript[(OverHat[J]), 2]))((Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]))
=(Subscript[(OverHat[J]), 1]^2)∓i (Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 2])+(Subscript[(OverHat[J]), 2]^2)
=((Subscript[(OverHat[J]), 1]^2)+(Subscript[(OverHat[J]), 2]^2))±i ((Subscript[(OverHat[J]), 1])(Subscript[(OverHat[J]), 2])-(Subscript[(OverHat[J]), 2])(Subscript[(OverHat[J]), 1]))
=(((OverHat[J])^2)-(Subscript[(OverHat[J]), 3]^2))∓(Subscript[(OverHat[J]), 3])
=((OverHat[J])^2)-(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±1).

(Subscript[(OverHat[J]), ∓]^(l+1))(Subscript[(OverHat[J]), ±]^(l+1))=(Subscript[(OverHat[J]), ∓]^l)(((OverHat[J])^2)-(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±1))(Subscript[(OverHat[J]), ±]^l)
=(Subscript[(OverHat[J]), ∓]^l)((OverHat[J])^2)(Subscript[(OverHat[J]), ±]^l)-(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±1)(Subscript[(OverHat[J]), ±]^l).

[((OverHat[J])^2),(Subscript[(OverHat[J]), ±])]=[((OverHat[J])^2),(Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2])]
=[((OverHat[J])^2),(Subscript[(OverHat[J]), 1])]±i[((OverHat[J])^2),(Subscript[(OverHat[J]), 2])]
=0±i 0=0.

((OverHat[J])^2)(Subscript[(OverHat[J]), ±]^l)=(Subscript[(OverHat[J]), ±]^l)((OverHat[J])^2).

(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±]^l)=(Subscript[(OverHat[J]), ±]^l)((Subscript[(OverHat[J]), 3])±l),

(Subscript[(OverHat[J]), 3])((Subscript[(OverHat[J]), 3])±1)(Subscript[(OverHat[J]), ±]^l)=(Subscript[(OverHat[J]), ±]^l)((Subscript[(OverHat[J]), 3])±l)((Subscript[(OverHat[J]), 3])±(l+1)).

(Subscript[(OverHat[J]), ∓]^(l+1))(Subscript[(OverHat[J]), ±]^(l+1))=(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)((OverHat[J])^2)-(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)((Subscript[(OverHat[J]), 3])±l)((Subscript[(OverHat[J]), 3])±(l+1))
=(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±l)((Subscript[(OverHat[J]), 3])±(l+1))).

Quantum Bootstrap

Problem Setup

((OverHat[J])^2)(Ket[j,m])=(Subscript[λ, j])(Ket[j,m]),
(Subscript[(OverHat[J]), 3])(Ket[j,m])=(Subscript[λ, m])(Ket[j,m]),

(Subscript[λ, j]) is the the eigenvalue of ((OverHat[J])^2) of the (Ket[j,m]) state,

(Subscript[λ, m]) is the the eigenvalue of (Subscript[(OverHat[J]), 3]) of the (Ket[j,m]) state.

General Principles

Eigen condition

(Bra[j,m])f(((OverHat[J])^2),(Subscript[(OverHat[J]), 3]))(OverHat[O]) (Ket[j,m])
=(Bra[j,m])(OverHat[O]) f(((OverHat[J])^2),(Subscript[(OverHat[J]), 3]))(Ket[j,m])
=f((Subscript[λ, j]),(Subscript[λ, m]))(Bra[j,m])(OverHat[O])(Ket[j,m]),

for any function f. In particular, it implies

Positivity constraint

Angular Momentum Quantization

[(Subscript[(OverHat[J]), 3]),(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))]=∓(l-l')(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l')),

Prove Eq. () using Eq. ().

[(Subscript[(OverHat[J]), 3]),(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))]=(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))-(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))(Subscript[(OverHat[J]), 3])
=(Subscript[(OverHat[J]), ∓]^l)((Subscript[(OverHat[J]), 3])∓l)(Subscript[(OverHat[J]), ±]^(l'))-(Subscript[(OverHat[J]), ∓]^l)((Subscript[(OverHat[J]), 3])∓l')(Subscript[(OverHat[J]), ±]^(l'))
=(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±]^(l'))+(Subscript[(OverHat[J]), ∓]^l)(∓l)(Subscript[(OverHat[J]), ±]^(l'))-(Subscript[(OverHat[J]), ∓]^l)(Subscript[J, 3])(Subscript[(OverHat[J]), ±]^(l'))-(Subscript[(OverHat[J]), ∓]^l)(∓l')(Subscript[(OverHat[J]), ±]^(l'))
=∓(l-l')(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))

(Bra[j,m])[(Subscript[(OverHat[J]), 3]),(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))](Ket[j,m])=∓(l-l')(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))(Ket[j,m]).

(l-l')(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))(Ket[j,m])=0.

If l≠l', we must have (Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))(Ket[j,m])=0.

If l=l', Eq. () is automatically satisfied, and there is no restriction on (Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))(Ket[j,m]). Its value remains to be determined, and can be defined as

(Subscript[A, j,m]^(±,l)) :=(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(Ket[j,m]).

(Subscript[A, j,m]^(±,l+1))=((Subscript[λ, j])-((Subscript[λ, m])±l)((Subscript[λ, m])±(l+1)))(Subscript[A, j,m]^(±,l)).

Derive Eq. () using Eq. ().

(Subscript[A, j,m]^(±,l+1))=(Bra[j,m])(Subscript[(OverHat[J]), ∓]^(l+1))(Subscript[(OverHat[J]), ±]^(l+1))(Ket[j,m])
=(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(((OverHat[J])^2)-((Subscript[(OverHat[J]), 3])±l)((Subscript[(OverHat[J]), 3])±(l+1)))(Ket[j,m])
=((Subscript[λ, j])-((Subscript[λ, m])±l)((Subscript[λ, m])±(l+1)))(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(Ket[j,m])
=((Subscript[λ, j])-((Subscript[λ, m])±l)((Subscript[λ, m])±(l+1)))(Subscript[A, j,m]^(±,l))

(Subscript[A, j,m]^(±,l))=(∏_(k=0)^(l-1))((Subscript[λ, j])-((Subscript[λ, m])±k)((Subscript[λ, m])±(k+1))).

(Subscript[A, j,m]^(±,l))=(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^l)(Ket[j,m])≥0,

(∏_(k=0)^(l-1))((Subscript[λ, j])-((Subscript[λ, m])±k)((Subscript[λ, m])±(k+1)))≥0.

Manipulate[Block[{pltrng={{-3,3},{0,14}},region},
SetAttributes[region,Listable];
region[λj==y_,λm==x_]:=Point[{x,y}];
region[Inequality[y1_,LessEqual,λj,Less,y2_],λm==x_]:={Line[Table[{x/.λj->y,y},{y,y1,y2,0.1}]],Point[{x/.λj->y1,y1}]};
region[Inequality[y1_,Less,λj,Less,y2_],λm==x_]:={Line[Table[{x/.λj->y,y},{y,y1,y2,0.1}]]};
region[Inequality[λj,Greater,y1_],λm==x_]:={Line[Table[{x/.λj->y,y},{y,y1,Max[Last@pltrng],0.1}]]};
region[Inequality[λj,Greater,y1_],Inequality[x1_,LessEqual,λm,LessEqual,x2_]]:=Block[{pts1,pts2},
{pts1,pts2}=Table[{#/.λj->y,y},{y,y1,Max[Last@pltrng],0.1}]&/@{x1,x2};
{{Opacity[0.2],Polygon[Join[pts1,Reverse[pts2]]]},
Line[pts1],Line[pts2]}];
Graphics[{AbsolutePointSize[5],ReplaceRepeated[Reduce[And@@Flatten@Table[{Product[λj-(λm-k)(λm-(k+1)),{k,0,l-1}]>=0,Product[λj-(λm+k)(λm+(k+1)),{k,0,l-1}]>=0},{l,1,lmax}],{λj,λm}],{And->region,Or->List}]},
GridLines->{Range[-3,3,1/2],#(#+1)&/@Range[0,3,1/2]},Frame->True,FrameLabel->{"\!\(λ\_m\)","\!\(λ\_j\)"},PlotRange->pltrng,PlotRangeClipping->True,PlotRangePadding->{{0.2,0.2},{0.5,0}},PlotLabel->Style[StringTemplate["\!\(l\_max=`1`\)"][lmax],14,Black],AspectRatio->1,ImageSize->180]],{lmax,Range[1,8,1],ImageSize->Small},Paneled->False]

For orbital angular momentum j takes integer values. 
For spin angular momentum j can also be half-integers.

The eigen equations in Eq. () become

((OverHat[J])^2)(Ket[j,m])=j(j+1)(Ket[j,m]),
(Subscript[(OverHat[J]), 3])(Ket[j,m])=m(Ket[j,m]).

The expectation value reads

(Bra[j,m])(Subscript[(OverHat[J]), ∓]^l)(Subscript[(OverHat[J]), ±]^(l'))(Ket[j,m])=({{{, ({{0, if l≠l'}, {(∏_(k=0)^(l-1))(j(j+1)-(m±k)(m±(k+1))), if l=l'}})}}).

Operator Representation

(Subscript[(OverHat[J]), 3])(Subscript[(OverHat[J]), ±])(Ket[j,m])=(Subscript[(OverHat[J]), ±])((Subscript[(OverHat[J]), 3])±1)(Ket[j,m])
=(m±1)(Subscript[(OverHat[J]), ±])(Ket[j,m])

(Subscript[(OverHat[J]), ±])(Ket[j,m])=(Subscript[c, j,m]^±)(Ket[j,m±1]).

(Bra[j,m])(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])(Ket[j,m])=j(j+1)-m(m±1).

(Bra[j,m])(Subscript[(OverHat[J]), ∓])(Subscript[(OverHat[J]), ±])(Ket[j,m])=((((Subscript[c, j,m]^±)))^2)(BraKet[j,m±1, j,m±1])=((((Subscript[c, j,m]^±)))^2).

(Subscript[c, j,m]^±)=(Sqrt[j(j+1)-m(m±1)]).

either raised from the lowest weight state,

(Ket[j,m])=((((((j-m)!)/((2j)!(j+m)!))))^(1/2))(Subscript[(OverHat[J]), +]^(j+m))(Ket[j,-j]),

or lowered from the highest weight state,

(Ket[j,m])=((((((j+m)!)/((2j)!(j-m)!))))^(1/2))(Subscript[(OverHat[J]), -]^(j-m))(Ket[j,j]).

((SubPlus[(OverHat[J])])/(Sqrt[2j]))(Ket[j,-j+n])=(Sqrt[n+1])(Ket[j,-j+n+1])+𝒪((j^(-1/2))),
((SubMinus[(OverHat[J])])/(Sqrt[2j]))(Ket[j,-j+n])=(Sqrt[n])(Ket[j,-j+n-1])+𝒪((j^(-1/2))).

(Ket[j,-j+n])→(Ket[n]),
(((2j))^(-1/2))(SubMinus[(OverHat[J])])->a, (((2j))^(-1/2))(SubPlus[(OverHat[J])])→(SuperDagger[a]),

Summary

(OverHat[J])×(OverHat[J])=i (OverHat[J]).

The total angular momentum operator

((OverHat[J])^2)=(Subscript[(OverHat[J]), 1]^2)+(Subscript[(OverHat[J]), 2]^2)+(Subscript[(OverHat[J]), 3]^2).

The raising and lowering operators

(Subscript[(OverHat[J]), ±])=(Subscript[(OverHat[J]), 1])±i (Subscript[(OverHat[J]), 2]).

((OverHat[J])^2)(Ket[j,m])=j(j+1)(Ket[j,m]),
(Subscript[(OverHat[J]), 3])(Ket[j,m])=m(Ket[j,m]),
(Subscript[(OverHat[J]), ±])(Ket[j,m])=(Sqrt[j(j+1)-m(m±1)])(Ket[j,m±1]),

j=0,(1/2),1,(3/2),…,
m=-j,-j+1,…,j-1,j.